# A conjecture about irreducible polynomials with integer coefficients

Let $$f\in\mathbb Z[X]$$, define $$\operatorname{P}^+(f)$$ as the number of primes $$>0$$ that $$f$$ assumes at distinct integral arguments.

Theorem: If $$f\in\mathbb Z[X]$$ is non constant and reducible of degree $$n$$, then $$\operatorname{P}^+(f)\leq n$$. And for all $$n$$ there are non constant reducible polynomials of degree n such that $$\operatorname{P}^+(f)=n$$.
[Acta Arith.,104.2 (2002) 117-127.]

Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $$>1$$ and wouldn't pass the test.

A conjecture related to the conjecture of Bunjakowsky states:

Conjecture: If $$f\in\mathbb Z[X]$$ is non constant and irreducible, then $$f(a)/d$$ assumes primes for an infinit number of distinct integral arguments $$a$$, where $$d$$ is the largest fixed divisor of $$f$$.

This makes me wonder if the following hypothesis is true:

$$f\in\mathbb Z[X]$$ of degree $$n>0$$ with coprime coefficients is irreducible, iff $$\;\operatorname{P}^+(d^{-1}\cdot f)> n$$ or $$\;\operatorname{P}^+(d^{-1}\cdot (-f))> n$$, where $$d$$ is the greatest fixed divisor of $$f$$.

$$\operatorname{P}^+$$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?

With a test program using the hypothesis on Eisenstein polynomials $$f$$ with random coefficients between $$-19$$ and $$19$$ and random degree between $$2$$ and $$5$$, testing both $$f$$ and $$-f$$, resulted in no miss in $$1,000,000$$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.

• Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(\mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = \pm p$ then $g(n) = \pm 1$ or $h(n) = \pm 1$. Write $g(X) = 1+g_2(X)\prod_{n \in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d \ne 1$ ? Jan 20, 2019 at 8:43
• @reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
– Lehs
Jan 20, 2019 at 9:18
• @JovanRadenkovic: These two polynomials are not over $\mathbb Z$. Or what do you mean?
– Lehs
Jan 7, 2023 at 16:30
• @Lehs, I meant $x\cdot(x^2-8x+17)$ and $x\cdot(x^2+17)$, respectively.
– user1115547
Jan 7, 2023 at 16:32
– Lehs
Jan 7, 2023 at 20:36

• The simplest case :

Let $$f \in \mathbb{Z}[X]$$. It is said irreducible iff $$f(X) = g(X)h(X) \implies g(X)=\pm1$$ or $$h(X) = \pm1$$. Let $$d = gcd(f(\mathbb{Z}))$$. Assume $$d=1$$ and $$| f(n)|$$ is prime more than $$2 \deg(f)$$ times.

If $$f(X) = g(X)h(X)$$ is reducible, then $$f(n) = \pm p$$ implies $$n$$ is a root of $$g(X)^2-1$$ or $$h(X)^2-1$$. But those polynomials are of degree $$2\deg(g), 2 \deg(f)-2\deg(g)$$, so they have at most $$2\deg(f)$$ roots. A contradiction. Whence $$f$$ is irreducible.

• To improve the bound $$2 \deg(f)$$, you need to split each case : $$g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$$ and factorize.

• Then look at the case $$d=2$$, you'll have more cases and some of them will probably allow more than $$\deg(f)$$ prime values.

• You are right and I have corrected the question.
– Lehs
Jan 23, 2019 at 9:09